Understanding boolean algebra and boolean axioms?

Question

I'm currently studying discrete mathematics and am having some difficulties with understanding boolean algebra. To be specific, I'm stuck on the following question:

Let A = {a, b} and list the four elements of the power set P(A). We consider
the operations + to be ∪, · to be ∩, and complement to be set complement.
Consider 1 to be A and 0 to be ∅.

 1. Explain why the description above defines a Boolean algebra
 2.  Find two elements x, y in P(A) such that xy = 0, x != 0 and y != 0.

Starting with the power set

P(A) = {∅, {a},{b},{a,b}}

How would I go about finding the elements of x & y to satisfy part two of the question using algebraic axioms? Also, for explaining how the above defines a boolean algebra, do you think it would suffice to simply mention how there are two binary operations and a set associated with the boolean algebra? Any help is appreciated, thanks!

Answer 1

Compare:$\begin{array}{c|cc}\cup & \varnothing & \{a\} & \{b\} & \{a, b\} \\ \hline \varnothing & \varnothing & \{a\} & \{b\} & \{a, b\} \\ \{a\} &\{a\} & \{a\} & \{a, b\} & \{a, b\} \\ \{b\} & \{b\} & \{a, b\} & \{b\} & \{a, b\} \\ \{a, b\} & \{a, b\} & \{a, b\} & \{a, b\} & \{a, b\} \end{array}\qquad\begin{array}{c|cc}{+} & 0 & X & X^\complement & 1 \\ \hline \varnothing & 0 & X & X^\complement & 1 \\ X & X & X & 1 & 1 \\ X^\complement & X^\complement & 1 & X^\complement & 1 \\ 1 & 1 & 1 & 1 & 1 \end{array}$

Do likewise for $\cap$ and $\cdot$

Answer 2

Following up on @bof's snide remark:

One way to find $x$ and $y$ is trial and error. Sometimes -- esp. in cases like this one where the number of possibilities is small -- that's not a bad way to start. In the course of trying-and-erroring, you're likely to notice some patterns that make this or that example almost work, and you can then more quickly home in on one that actually will work.

Since this is a HW problem, perhaps you were hoping that there was some clever way to figure out the sets $x$ and $y$ using formulas, but I know of no such thing. You CAN figure out $x$ and $y$ by reasoning carefully, but trial end error should be just about as fast. Anyhow, since it's HW, I'm not going to tell you the answer, and merely leave it at this hint, along with the associated advice that sometimes trying small experiments can lead to insights.